Question

How many positive integers a less than 100 have a corresponding integer b divisible by 3 such that the roots of x^2-ax+b=0 are consecutive positive integers?

Answer 1

There are 24 such numbers

5, 7, 11, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 61, 65, 67, 71, 79, 83, 85, 91, 95

Explanation:

Solving, we have

(x-y)(x-(y-1))=x^2-(2×y-1)x+y^2-y=x^2-ax+b comparison shows that

(2y-1)=a and y^2-y =b or

y(y-1)=b

This gives the following qualifying numbers

5, 7, 11, 17, 19, 23, 25, 29, 31, 35, 37, 41, 43, 47, 49, 61, 65, 67, 71, 79, 83, 85, 91, 95